For $a \neq 0$ and $b \neq 0$,if the real valued function $f(x) = \frac{\sqrt[5]{a(625+x)} - 5}{\sqrt[4]{625+bx} - 5}$ is continuous at $x = 0$,then $f(0) =$

Let $f: R \rightarrow R$ be a function given by $f(x) = \begin{cases} \frac{1-\cos 2x}{x^2} & , x < 0 \\ \alpha & , x = 0 \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $\alpha^2 + \beta^2$ is equal to:

Let $f(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \ge 0 \end{cases}$,then for all values of $x$

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is

The number of points at which the function $f(x) = \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}{6-2\sqrt{2+|x|}}$ is discontinuous in $(-\infty, \infty)$ is

If $f(x) = \frac{x^2-10x+25}{x^2-7x+10}$ and $f$ is continuous at $x=5$,then $f(5)$ is equal to